# Homework Help: Epsilon-limit proof for real number sequences

1. Jan 18, 2010

### kingwinner

1. The problem statement, all variables and given/known data
Let {an} be a sequence of real numbers.
Suppose an->L as n->∞. Prove that [(a1+a2+...+an)/n] ->L as n->∞.

2. Relevant equations
N/A

3. The attempt at a solution
By definition:
an->L iff
for all ε>0, there exists an integer N such that n≥N => |an - L|< ε.

Given ε>0.
I start with |[(a1+a2+...+an)/n] - L|. But I cannot think of a way to link this with |an - L| so I have no idea how to continue.

Any help is appreciated! :)

2. Jan 18, 2010

### kingwinner

I think we may have to use the triangle inequality, but still I have no idea how to get into the exact the same form as |an - L|.

Could someone help me, please?

Last edited: Jan 18, 2010
3. Jan 18, 2010

### kingwinner

I don't get this part.

In the hypothesis, we suppose an->L, so:
for all εa>0, there exists an integer N such that n≥N => |an - L|< εa.

So for example:
Given ε>0. We can find N s.t. n≥N => |an-L|< ε/4

But how come you have |(a1+...+an) / n|?

Thanks for your help! :)

4. Jan 18, 2010

### LCKurtz

It isn't. I must have been in a fog when I posted that and I have deleted it. Be back a little later.

5. Jan 18, 2010

### LCKurtz

Given $\varepsilon > 0$ pick N such that if n > N then $|a_n - L| < \varepsilon/ 4$. Now if n > N

$$\left|\frac{a_1+...+a_n}{n}-L\right| = \left|\frac{a_1+...+a_n}{n}-\frac {nL}{n}\right|= \left|\frac{(a_1-L)+...+(a_n-L)} n\right|$$

$$= \left| \frac{(a_1-L)+...+(a_N-L)}{n}+\frac{(a_{N+1}-L)+...+(a_n-L)}{n}\right|$$

Now use the triangle inequality. Since N is fixed you can make the first term small in n is large enough. Then work on the second term to make it small. Sorry for the false start earlier.

6. Jan 18, 2010

### kingwinner

1) hmm...why did you choose ε/4 in $|a_n - L| < \varepsilon/ 4$??

2) At the end, we have to show that:
for all ε>0, there exists an integer M such that
n≥M => |[(a1+a2+...+an)/n] - L|< ε.
But how can we find such an M?

Thanks for your help! :)

Last edited: Jan 18, 2010
7. Jan 18, 2010

### LCKurtz

Because I anticipate adding a couple of small terms together and having the sum come out less than $\varepsilon$.

Have you done the triangle inequality on this yet, so you have two terms to make small? If you can say there exists an N1 such that n > N1 makes the first term small and an N2 such that n > N2 makes the second term small, then if you take Nmax = max{N1,N2,N}, wouldn't that do it? Of course, you have to fill in the arguments why you can pick these N's.

8. Jan 18, 2010

### kingwinner

So after using the triangle inequality (which will break it into 2 terms), for the first term I have to find N1 such that n≥N1 => |[(a1 -L)+...(aN -L)]/n| < ε/2

To find this N1, I think I have to relate |[(a1 -L)+...(aN -L)]/n| to |an - L|, but how is this possible?? I can't think of a way of doing it...

Thanks!

9. Jan 18, 2010

### LCKurtz

N is a fixed, finite, number. There are finitely many terms in the numerator and you have n in the denominator. Just pick n big enough...

10. Jan 18, 2010

### kingwinner

I get the intuitive idea, but just like every epsilon-limit proof of this type, we have to actually find a specific N1 which works, right? And I'm stuck on this step...

11. Jan 18, 2010

### LCKurtz

So solve your inequality |[(a1 -L)+...(aN -L)]/n| < ε/2 for n to see how large it must be.

12. Jan 18, 2010

### kingwinner

OK. I think it is n> 2 |(a1-L)+...+(aN-L)| / ε.
So I think we can take N1 = 2 |(a1-L)+...+(aN-L)| / ε, is this correct?
But on top of N depending on ε, N also depends on the terms in the sequence, is this OK? (i.e. is this allowed?)

13. Jan 18, 2010

### LCKurtz

Everything depends on the terms of the sequence. You wouldn't expect the same N to work for all sequences, would you?

Remember you might have to take N1 slightly larger than that because that might not be an integer.

14. Jan 18, 2010

### kingwinner

So I guess I will take N1 to be the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.

Any hints for the second term? (sorry I'm asking it but I really have no clue...)

|[aN+1 - L +...+ (an-1-L) +(an-L)] / n|

Here there are terms like an, an-1, etc. The "n" pops up everyhwere, so I don't think I can possibly solve for n in the inequality just like I did before.

15. Jan 18, 2010

### r.a.c.

The sequence an converges therefore it is bounded. So every an<M where M is some number. So the sum becomes less that (M-L) ...n-N+1 times over n and from there go on.

16. Jan 18, 2010

### kingwinner

|[aN+1 - L +...+ (an-1-L) +(an-L)] / n|
≤|(n-N)(M-L)/n|

I think it should be n-N instead of n-N+1?
n-(N+1)+1 = n-N

17. Jan 18, 2010

### LCKurtz

Somewhere in this proof you need to use the fact that the original sequence converges to L. Remember at the beginning where you had that if n > N,

$|a_n - L| < \varepsilon/ 4$

Use that now.

18. Jan 19, 2010

### kingwinner

Do I have to use the fact that any convergence sequence is bounded?

But how should I handle the terms like (an-1-L) in the numerator?

thanks.

19. Jan 19, 2010

### LCKurtz

I fear you are getting lost in the subscripts. In the expression:

|[aN+1 - L +...+ (an-1-L) +(an-L)] / n|

all the subscripts are greater than N. So every one of them satisfies the inequality:

$|a_k - L| < \varepsilon/ 4$

for whatever value from N+1 to n that k takes. Use that.

20. Jan 19, 2010

### kingwinner

So should I break |[(aN+1 - L) +...+ (an-1-L) +(an-L)] / n| using the triangle inequality into n-N terms?
|[(aN+1 - L) +...+ (an-1-L) +(an-L)] / n| <= (1/n)[|aN+1 - L| + ...+ |an-1-L| + |an-L|] ?

But I don't think it will work...

21. Jan 19, 2010

### LCKurtz

You won't know until you try. How big are the terms? How many terms are there?

22. Jan 19, 2010

### kingwinner

There are n-N terms.

|[(aN+1 - L) +...+ (an-1-L) +(an-L)] / n| <= (1/n)[|aN+1 - L| + ...+ |an-1-L| + |an-L|] < (1/n)(n-N)ε/4

I can't solve for n here as I did for the first part, so I am not sure what to do next...

23. Jan 19, 2010

### LCKurtz

How big can

$$\left(\frac {n-N}{n}\right)\left(\frac\varepsilon 4\right)$$

be?

24. Jan 19, 2010

### kingwinner

It goes to 0 as n->infinity.

But how can we find a specific N2 which works (just like before when we found N1) in the epsilon-limit proof?

25. Jan 19, 2010

### LCKurtz

You aren't looking for another N and we aren't talking about n -> infinity. All you are trying to do is show this sum is small. You have that it is smaller than

$$\left(\frac {n-N}{n}\right)\left(\frac\varepsilon 4\right)$$

I will ask you again. Is that small? Look at it and think about it.

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